Problem: $\int e^{2x}\,dx\,= $ $+~C$
Explanation: If we let $ {u=2x}$, then ${du=2 \, dx}$ and $ dx=\dfrac{du}{2}}$. Substituting gives us: $\begin{aligned} \int e^{{2x}}\, dx}\,&= \int e^{ u}\, \dfrac{du}{2}}\,\\\\\\ &= \int e^u\cdot \dfrac12\, du\\\\\\ &=\dfrac12 \int e^u\, du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&=\dfrac12 \int e^u\, du~\\\\\\ &=\dfrac12e^u+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 2 e u + C = 1 2 e 2 x + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac12e^u+C\\\\\\\ &=\dfrac12e^{ 2x}+C\end{aligned} The answer: $\int e^{2x}\,dx\,= \dfrac12e^{ 2x}+C$